I'm so stuck on this. I have no clue where to start.

sec (tan^-1 4)

(the tan is an inverse with the -1, 4 is the number)

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- Apr 6th 2009, 12:25 PMtiarInverse trignometry; sec and tan
I'm so stuck on this. I have no clue where to start.

sec (tan^-1 4)

(the tan is an inverse with the -1, 4 is the number)

- Apr 6th 2009, 12:34 PMstapel
Does this mean that you're not familiar with the meanings of the inverse trigonometric functions?

In case you're exaggerating a bit...

By definition, arctan(4) = A means tan(A) = 4 = 4/1.

Draw a right triangle. Label the non-right base angle as A. Label the "opposite" and "adjacent" sides as 4 and 1, respectively.

Use the Pythagorean Theorem to find the length of the hypotenuse.

Read the value of the sine from the triangle. (Wink) - Apr 6th 2009, 12:39 PMtiar
No, I'm not very familiar with them at all.

Okay, so the hyp. of the right triangle would be $\displaystyle \sqrt{17}$ thus sine would be $\displaystyle \frac{4}{\sqrt{17}}$?

And I need to know the quadrant it is in to know if it is negative or positive, correct?

I'm confused on how to do that. - Apr 6th 2009, 12:40 PMicemanfan
You are looking for the secant of the angle whose tangent is 4. Remember that in a right triangle, $\displaystyle x^2 + y^2 = r^2$. Imagine a right triangle in which one leg (y) is four times as long as the other (x). This is indicated by the tangent (y/x) = 4. Hence,

$\displaystyle \frac{y}{x} = 4$

$\displaystyle y = 4x$

$\displaystyle x^2 + (4x)^2 = r^2$

$\displaystyle x^2 + 16x^2 = r^2$

$\displaystyle 17x^2 = r^2$

$\displaystyle 17 = \frac{r^2}{x^2}$

$\displaystyle \sqrt{17} = \frac{r}{x}$

Since the secant is r/x, it is equal to $\displaystyle \sqrt{17}$. - Apr 6th 2009, 12:44 PMicemanfan